I buried four fishermen up to their necks in the sand on the beach at low tide for keeping their fishing spot a secret from me. I put a hat on each of their heads and told them that one of them must shout out the correct color of their own hat or they will all be drowned by the incoming tide. I give them 10 minutes to do this. Fisherman A and B can only see the sand dune I erected. Fisherman C can see that fisherman B has a white hat on. Fisherman D can see that C has a black hat, and B has a white hat. The fisherman have been told that there are four hats, two white and two black, so they know that they must have either a white or a black hat on. who shouts out the color of their hat and how do they know?
Fisherman C shouts out.
Fishermen A and B are in the same situation - they have no information to help them determine their hat colour so they can’t answer. C and D realise this.
Fisherman D can see both B and C’s hats. If B and C had the same colour hat then this would let D know that he must have the other colour.
When the time is nearly up, or maybe before, C realises that D isn’t going to answer because he can’t. C realises that his hat must be different to B’s otherwise D would have answered. C therefore concludes that he has a black hat because he can see B’s white on
Friday, April 1, 2011
Who Will Tell The Color of His Hat/halos.. First
Four angels sat on the Christmas tree amidst other ornaments. Two had blue halos and two – yellow. However, none of them could see above his head. Angel A sat on the top branch and could see the angels B and C, who sat below him. Angel B, could see angel C who sat on the lower branch. And angel D stood at the base of the tree obscured from view by a thicket of branches, so no one could see him and he could not see anyone either.
Which one of them could be the first to guess the color of his halo and speak it out loud for all other angels to hear?
Case 1: Angels B and C have same color(either both blue or yellow) Then A can be sure of his color.
Case 2: Angles B and C have different colors.
B after waiting for some time will realise that A has not spoken out the color yet ,hence B and C must have different colors. B will look at C's color and announce his color opposite to C's color.
Which one of them could be the first to guess the color of his halo and speak it out loud for all other angels to hear?
Case 1: Angels B and C have same color(either both blue or yellow) Then A can be sure of his color.
Case 2: Angles B and C have different colors.
B after waiting for some time will realise that A has not spoken out the color yet ,hence B and C must have different colors. B will look at C's color and announce his color opposite to C's color.
Wednesday, March 30, 2011
Weigh System With More Explaination..How Many Weight We Needs..???
Puzzle: You have a common balance with two sides. You can only keep weighs on one side. You have to weigh items from 1 to 1000 weight units with integral weights only.
What is the minimum number of weights required and what will be their weights?
Eg. With 2 weights one of 1 weight unit and other of 3 weight unit you can measure 1, 3 and 4 only.
Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow. You can provide your answer in comments.
Solution: This puzzle is an easy puzzle as to make any odd number you need all other even numbers and one 1.
So only one odd number is needed to make all odd numbers. Now you are left with all even numbers. All even numbers are base 2 and thus can be though of as binary representation. To make 1000 you need at most 10 bits(1111101000). You can make any number below 1000 from these bits.
So the answer is 10 weights required. 1, 2, 4, 8,16, 32, 64, 128, 256, 512.
Puzzle: This is part 2 of puzzle posted Above. The puzzle is that you have to weigh 1 to 1000 using a common balance scale with minimum number of weighs. This time you can put wights on both sides.
You can check Part 1 of this puzzle How many weights? Puzzle Part 1.
Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow with winner's name. You can provide your answer in comments.
Solution: The answer is series of base 3 i.e. 1, 3, 9, 27, 81, 243 and 729.
We can start from smallest number 1. To measure 1 we definitely need 1.
To measure 2 we need to look for highest number possible. We can measure 2 by using 3(3-1 = 2). So we need 1 and 3.
Now to measure 4 we can use 1 and 3 both (1+ 3). To measure 5 again we will look for highest possible number. 9 is highest possible number (9-4 = 5). We can measure up to 13 using these weights.
Continuing the same way we can deduce that a series of base 3 is required, thus 1, 3, 9, 27, 81, 243, 729.
Puzzle: This is the 3rd part of weighing puzzles. You can check Part 2 and Part 1.
You have to weigh 125 packets of sugar with first one weighing 1 lb, second 2 lb, third 3 lb and so on.You can only use one pan of the common balance for measurement for weighing sugar, the other pan has to be used for weights.
The weights have to put on to the balance and it takes time. Time taken to move weights is equal to the number of weights moved.
Eg. So if you are weighing 5 lb using 3lb and 2 lb weight then you have to lift both these weights one by one and thus it will take 2 unit time. If you have to measure only 3lbs it will take only 1 unit time. Now if we have to measure sugar packets of weights 1,3,4 using weights 1 and 3 only. For this first we measure 1 lb, with 1 unit of time, we place 3 lb along with 1 lb and measure 4kg with again 1 unit of time, and finally we move 1kg out of pan to measure 3kg in 1 unit of time. So in 3 units of time we could measure 1,3 and 4lb using weights 1 and 3 only.
Now you have to make sugar packets of all weights from 1 to 125 in minimum time, in other words in minimum movement of weights. The question here is to find out the minimum number of weighs needed and the weight of each the weights used and the strategy to be followed for the creation of 125 packets of sugar.
Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow with winner's name. You can provide your answer in comments.
Source: Classic Puzzles
Solution: Amazing explanation provided by Kasturi.
You will need the weights 1, 2, 4, 8, 16, 32 and 64 i.e. 7 weights. You can measure 125 packets in 125 movements. The strategy to be followed is this:
Put 1: measure 1, weights in pan 1
To measure further you need to use a new weight.
Put 2: measure 3, weights in pan 1,2
Remove 1: measure 2, weights in pan 2
Thus, by adding the next smallest weight to the pan i.e. 2, we can measure all weights below 4 in total 3 movements.
To measure further you need to use a new weight.
Put 4: measure 6, weights in pan 2,4
Remove 2: measure 4, weights in pan 4
Put 1: measure 5, weights in pan 1,4
Put 2: measure 7, weights in pan 1,2,4
Thus, by adding the next smallest weight to the pan i.e. 4, we can measure all weights below 8 in total 7 movements.
Following the same strategy, whenever we have measured all weights below n, add n and we can measure measure all the weights below 2n in 2n-1 movements.
Eventually, we can measure 125 packets in 125 movements.
What is the minimum number of weights required and what will be their weights?
Eg. With 2 weights one of 1 weight unit and other of 3 weight unit you can measure 1, 3 and 4 only.
Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow. You can provide your answer in comments.
Solution: This puzzle is an easy puzzle as to make any odd number you need all other even numbers and one 1.
So only one odd number is needed to make all odd numbers. Now you are left with all even numbers. All even numbers are base 2 and thus can be though of as binary representation. To make 1000 you need at most 10 bits(1111101000). You can make any number below 1000 from these bits.
So the answer is 10 weights required. 1, 2, 4, 8,16, 32, 64, 128, 256, 512.
Puzzle: This is part 2 of puzzle posted Above. The puzzle is that you have to weigh 1 to 1000 using a common balance scale with minimum number of weighs. This time you can put wights on both sides.
You can check Part 1 of this puzzle How many weights? Puzzle Part 1.
Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow with winner's name. You can provide your answer in comments.
Solution: The answer is series of base 3 i.e. 1, 3, 9, 27, 81, 243 and 729.
We can start from smallest number 1. To measure 1 we definitely need 1.
To measure 2 we need to look for highest number possible. We can measure 2 by using 3(3-1 = 2). So we need 1 and 3.
Now to measure 4 we can use 1 and 3 both (1+ 3). To measure 5 again we will look for highest possible number. 9 is highest possible number (9-4 = 5). We can measure up to 13 using these weights.
Continuing the same way we can deduce that a series of base 3 is required, thus 1, 3, 9, 27, 81, 243, 729.
Puzzle: This is the 3rd part of weighing puzzles. You can check Part 2 and Part 1.
You have to weigh 125 packets of sugar with first one weighing 1 lb, second 2 lb, third 3 lb and so on.You can only use one pan of the common balance for measurement for weighing sugar, the other pan has to be used for weights.
The weights have to put on to the balance and it takes time. Time taken to move weights is equal to the number of weights moved.
Eg. So if you are weighing 5 lb using 3lb and 2 lb weight then you have to lift both these weights one by one and thus it will take 2 unit time. If you have to measure only 3lbs it will take only 1 unit time. Now if we have to measure sugar packets of weights 1,3,4 using weights 1 and 3 only. For this first we measure 1 lb, with 1 unit of time, we place 3 lb along with 1 lb and measure 4kg with again 1 unit of time, and finally we move 1kg out of pan to measure 3kg in 1 unit of time. So in 3 units of time we could measure 1,3 and 4lb using weights 1 and 3 only.
Now you have to make sugar packets of all weights from 1 to 125 in minimum time, in other words in minimum movement of weights. The question here is to find out the minimum number of weighs needed and the weight of each the weights used and the strategy to be followed for the creation of 125 packets of sugar.
Note: Don't forget to visit us again. Answer to the puzzle will be posted tomorrow with winner's name. You can provide your answer in comments.
Source: Classic Puzzles
Solution: Amazing explanation provided by Kasturi.
You will need the weights 1, 2, 4, 8, 16, 32 and 64 i.e. 7 weights. You can measure 125 packets in 125 movements. The strategy to be followed is this:
Put 1: measure 1, weights in pan 1
To measure further you need to use a new weight.
Put 2: measure 3, weights in pan 1,2
Remove 1: measure 2, weights in pan 2
Thus, by adding the next smallest weight to the pan i.e. 2, we can measure all weights below 4 in total 3 movements.
To measure further you need to use a new weight.
Put 4: measure 6, weights in pan 2,4
Remove 2: measure 4, weights in pan 4
Put 1: measure 5, weights in pan 1,4
Put 2: measure 7, weights in pan 1,2,4
Thus, by adding the next smallest weight to the pan i.e. 4, we can measure all weights below 8 in total 7 movements.
Following the same strategy, whenever we have measured all weights below n, add n and we can measure measure all the weights below 2n in 2n-1 movements.
Eventually, we can measure 125 packets in 125 movements.
Tiger & Sheep Puzzle..Will Sheep Survive or Not
Hundred tigers and one sheep are put on a magic island that only has grass. Tigers can live on grass, but they would rather eat sheep. Its a Magic Iceland because if a Tiger eats the Sheep then it will become a sheep itself (and hence can be eaten up by another tiger).
Tigers don’t mind being a sheep, but they would never want themselves to be eaten up. All tigers are intelligent and they want to survive. They however, don’t care of survival of others.
Will the sheep survive or will it be eaten up?
Solution:
This problem and the problem of pirates belong to the same family of Puzzles, where the puzzle is solved by simplification. Lets Consider the case when there are less Tigers
If there is 1 tiger, then he will eat the sheep because he does not need to worry about being eaten. Sheep will Not survive.
If there are 2 tigers, Both of them knows that if he eats the Sheep, the other tiger will eat him. So, The Sheep will Survive.
If there are 3 tigers, then they each of them knows that if he eats up the Sheep, then Iceland will be left with 1 sheep and 2 Tigers and as shown in the previous case, the Sheep will survive. Hence each tiger will try to eat up the sheep. The sheep will Not Survive.
If there are 4 Tigers, then the sheep will Survive.
And so on….
So, If there are even number of tigers the sheep will Survive, else it will die. Hence, if there are 100 tigers the sheep will Survive.
Tigers don’t mind being a sheep, but they would never want themselves to be eaten up. All tigers are intelligent and they want to survive. They however, don’t care of survival of others.
Will the sheep survive or will it be eaten up?
Solution:
This problem and the problem of pirates belong to the same family of Puzzles, where the puzzle is solved by simplification. Lets Consider the case when there are less Tigers
If there is 1 tiger, then he will eat the sheep because he does not need to worry about being eaten. Sheep will Not survive.
If there are 2 tigers, Both of them knows that if he eats the Sheep, the other tiger will eat him. So, The Sheep will Survive.
If there are 3 tigers, then they each of them knows that if he eats up the Sheep, then Iceland will be left with 1 sheep and 2 Tigers and as shown in the previous case, the Sheep will survive. Hence each tiger will try to eat up the sheep. The sheep will Not Survive.
If there are 4 Tigers, then the sheep will Survive.
And so on….
So, If there are even number of tigers the sheep will Survive, else it will die. Hence, if there are 100 tigers the sheep will Survive.
A Fox, A Sheep, and A Sack of Hay
A farmer is travelling with a fox, a sheep and a small sack of hay. He comes to a river with a small boat in it. The boat can only support the farmer and one other animal/item. If the farmer leaves the fox alone with the sheep, the fox will eat the sheep. And if the farmer leaves the sheep alone with the hay, the sheep will eat the hay.
How can the farmer get all three as well as himself safely across the river?
Answer
1. The farmer takes the sheep across the river, then returns back.
2. The farmer takes the fox across the river.
3. The farmer takes the sheep back to the first side of the river.
4. The farmer leaves the sheep back on the first side of the river, and takes the hay to the other side.
5. The farmer returns to the first side of the river.
6. The farmer brings the sheep back to the second side.
How can the farmer get all three as well as himself safely across the river?
Answer
1. The farmer takes the sheep across the river, then returns back.
2. The farmer takes the fox across the river.
3. The farmer takes the sheep back to the first side of the river.
4. The farmer leaves the sheep back on the first side of the river, and takes the hay to the other side.
5. The farmer returns to the first side of the river.
6. The farmer brings the sheep back to the second side.
Pills Puzzle
A person was prescribed to take two pills (tablets), one each, from the two bottles viz. Bottle A and Bottle B, daily. The tablets are exactly look alike.
The medicines have to be taken exactly one tablet from each bottle, neither less nor more, else the medicines will not be effective.
One fine day, the patient popped out one tablet from Bottle A, but while taking the tablet from bottle B, by mistake, two tablets spilled over. Now he has three tablets in his hand, and he can't put back the extra tablet to Bottle B as all the tablets are identical in looks.
He has to ensure that he takes exactly one tablet from each of the bottle and at the same time he must avoid any wastage of the medicine.
Constraints:
1. Both bottles have equal number of tablets, say 30.
2. Tablets from both the bottles look exactly identical.
3. Medicine is very costly, so any kind of wastage is not affordable.
Problem Statement: How would you ensure that, in the above situation, you take exactly one tablet from each bottle, at the same time ensuring no wastage of the medicine.
Answer
- Take (1) Pill A from the bottle and add it to the 3 unknown pills. You now have (2) Pill A and (2) Pill B in your pile.
- Take each of the 4 pills and cut them in half.
- For each pill, put one of the halves in a pile on the right and one of the halves in a pile on the left.
- Each pile now contains 2 halves of Pill A and 2 halves of Pill B, which is the same as (1) Pill A and (1) Pill B in each pile.
Tic Tac Toi Who Will Wind the Game
Who will win in this game of Tic-Tac-Toe (Refer picture), and what was the last move played? No additional information is available apart from the picture.
Answer
see the game fron the begining.firsst chance will be of 0,which would be placed at [1,2].Now x would be placed at [1,1].Now the 0 would be placed at [3,1].At this point x has two choice ,either at position [2,1] or [3,3].consider x is placed at [3,3].but if it is placed at [3,3] then 0 would be placed at middle ie [2,2].but it is not shown in the fig.then we have 2nd option i.e [2,1].now 0 would be placed at [3,2],which is blocked by x at[3,3].now the last turn will be of 0 which iss placed at [2,2].hence 0 will win the match.
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